第6章31题

// 6_31: 已知,由单链表表示的线性表中,含有三类字符的数据元素(如:字母字符,数字字符和其他字符),

// 是编写算法构造三个以循环链表表示的线性表,使每个表中只含同一类的字符,

// 且利用原表中的节点空间作为这三个表的结点空间,头节点可另辟空间。

//

void Proceed(node* head, node** a, node** b, node** c)

{

    //使用a,b,c作为传出指针

 

    int i = 0;

 

    node* list[3] = {*a, *b, *c};

 

    node* temp = head;

 

    if (temp == NULL)

    {

        printf("原列表为空!n");

        return;

    }

 

    for (i = 0; i < 3; i++)

    {

        if (list[i] == NULL)

        {

            list[i] = (node*)malloc(sizeof(node));

 

            if (list[i] == NULL)

            {

                printf("内存分配出错!n");

                return;

            }

        }

 

        list[i]->data = ”;            //置头节点标记

        list[i]->pNext = NULL;

    }//为a、b、c赋初值

 

    node* s[3];

 

    for (i = 0; i < 3; i++)

    {

        s[i] = list[i];

    }//三个链表的临时指针

 

    while (temp != NULL)

    {

        i = 0;

 

        if (temp->data >= ‘0’ && temp->data <= ‘9’)

        {

            i = 1;

        }

        else if (tolower(temp->data) >= ‘a’ && tolower(temp->data) <= ‘z’)

        {

            i = 2;

        }

 

        s[i]->pNext = temp;

 

        s[i] = temp;

 

        temp = temp->pNext;

    }

 

    for (i = 0; i < 3; i++)

    {

        s[i]->pNext = list[i];

    }

 

 

    *a = list[0];

    *b = list[1];

    *c = list[2];

 

    return;

}

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9 responses to this post.

  1. Posted by 陌陌 on 2006年03月30日 at 10:56

    6-17完整程序#include "stdio.h"#define k 10struct node{ int num;};int main(){ struct node a[20]; struct node temp; int i,j; for(i=0;i<20;i++)  a[i].num=0; for(i=0;i<10;i++)  a[i].num=i+1; for(i=0;i<20;i++)  printf("%d ",a[i].num); printf("\n"); for(j=0;j<k;j++){  i=9;  do{   temp=a[i+j+1];   a[i+j+1]=a[i+j];   a[i+j]=temp;   i–;  }while(i!=-1);  for(i=0;i<20;i++)   printf("%d ",a[i].num);  printf("\n"); } for(i=0;i<20;i++)  printf("%d ",a[i].num); return 0;}
     
     

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  2. Posted by 陌陌 on 2006年03月30日 at 10:57

    6-21数组法完整程序#include "stdio.h"#define N 10struct node{ int num;};int main(){ struct node a[N],temp; int i,j; for(i=0;i<N;i++)  a[i].num=i+1; for(i=0;i<N;i++)  printf("%d ",a[i].num); printf("\n"); for(i=0,j=N-1;i<=((N-1)/2),j>=((N+1)/2);i++,j–){
      temp=a[i];  a[i]=a[j];  a[j]=temp; } for(i=0;i<N;i++)  printf("%d ",a[i].num); return 0;
    }

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  3. Posted by 陌陌 on 2006年03月30日 at 10:58

    6-21链表法完整程序#include "stdio.h"#include "stdlib.h"#define N 10struct node{ int num; struct node * pNext;};int main(){ int i=0; struct node *head, *last; struct node * p1,* p2,* p3; head=(struct node *)malloc(sizeof(struct node)); head->pNext=NULL; head->num=i+1; i++; p1=head; while(i<N) {  p2=(struct node *)malloc(sizeof(struct node));  p1->pNext=p2;  p1=p1->pNext;  p1->num=i+1;  p1->pNext=NULL;  i++; } p1=head; while(p1!=NULL) {  printf("% d",p1->num);  last=p1;  p1=p1->pNext; } printf("\n"); p1=head; p2=p1->pNext; p3=p2->pNext; p1->pNext=NULL; while(1){ p2->pNext=p1; p1=p3->pNext; if(p1==NULL) {p3->pNext=p2; break;} p3->pNext=p2; p2=p1->pNext; if(p2==NULL) {p1->pNext=p3; break;} p1->pNext=p3; p3=p2->pNext; if(p3==NULL) {p2->pNext=p1; break;} } p1=last; while(p1!=NULL) {  printf("% d",p1->num);  p1=p1->pNext; } return 0;
     
     }
     

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  4. Posted by 陌陌 on 2006年03月30日 at 10:59

    6-27完整程序#include "stdio.h"#include "stdlib.h"#define N 10struct node{ int num; struct node * pNext;};int main(){ struct node * ha,* hb,*p1,*p2,*p3,*p4,*head; int i=0; p1=(struct node *)malloc(sizeof(struct node)); p2=(struct node *)malloc(sizeof(struct node)); ha=p1; hb=p2; p1->num=i+1; i++; p2->num=i+1; i++; while(i<N) {  p3=(struct node *)malloc(sizeof(struct node));  p4=(struct node *)malloc(sizeof(struct node));  p1->pNext=p3;  p3->num=i+1;  i++;  p1=p1->pNext;  p1->pNext=NULL;  p2->pNext=p4;  p4->num=i+1;  i++;  p2=p2->pNext;  p2->pNext=NULL; } p1=ha; p2=hb; head=NULL; while(p1->pNext!=NULL&&p2->pNext!=NULL) {  if(head==NULL)   if(p1->num>p2->num)    head=p2;   else    head=p1;  if(p1->num>p2->num){   p3=p2->pNext;   p2->pNext=p1;   p2=p3;  }  else if(p1->num<=p2->num){   p3=p1->pNext;   p1->pNext=p2;   p1=p3;  } } p1=head; while(p1) {  printf("% d",p1->num);  p1=p1->pNext; } return 0;}
     
     

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  5. Posted by 陌陌 on 2006年03月30日 at 10:59

    6-31完整程序#include "stdio.h"#include "conio.h"#include "stdlib.h"struct node{ char a; struct node * pNext;};int main(){ struct node * head=NULL,*p,*p1; p=(struct node *)malloc(sizeof(struct node)); head=p; head->pNext=NULL; p1=head; printf("Please input a string.(end with space)\n"); char b; b=getche(); head->a=b; while(b!=’ ‘){  b=getche();  p=(struct node *)malloc(sizeof(struct node));  p->a=b;  p1->pNext=p;  p1=p;  p1->pNext=NULL; } struct node * h1,*h2,*h3,*p2,*p3,*p4,*p5; p=(struct node *)malloc(sizeof(struct node)); h1=p; p1=h1; h1->a=’ ‘; p=(struct node *)malloc(sizeof(struct node)); h2=p; p2=h2; h2->a=’ ‘; p=(struct node *)malloc(sizeof(struct node)); h3=p; p3=h3; h3->a=’ ‘; p4=head; while(p4){  if((p4->a>=65&&p4->a<=90)||(p4->a>=97&&p4->a<=122)){   p5=p4->pNext;   p1->pNext=p4;   p4->pNext=h1;   p1=p4;   p4=p5;  }  else if(p4->a>=48&&p4->a<=57){   p5=p4->pNext;   p2->pNext=p4;   p4->pNext=h2;   p2=p4;   p4=p5;  }  else{   p5=p4->pNext;   p3->pNext=p4;   p4->pNext =h3;   p3=p4;   p4=p5;  } } printf("devise as:\n"); p1=h1; do{  p1=p1->pNext;  printf("%c",p1->a); }while(p1->a!=’ ‘); printf("\n"); p1=h2; do{  p1=p1->pNext;  printf("%c",p1->a); }while(p1->a!=’ ‘); printf("\n"); p1=h3; do{  p1=p1->pNext;  printf("%c",p1->a); }while(p1->a!=’ ‘); printf("\n"); return 0;}
     

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  6. Posted by my-space on 2006年04月1日 at 14:47

    #include<iostream.h>#include<stdio.h>#include<string.h>#include<iomanip.h>struct note2{ char ch[10]; note2 *next;};
    void main()         {    note2 *head=NULL;     //head为头指针    note2 *end=NULL; int last=0;    cout<<"输入字符串如e1,e2,….e9并以over作为终止符"<<endl; char num[10]; note2 *ip,*ir,*pnext,*phead,*iq;  /*在连表反转时:ir用于遍历节点     pnext指向ir指向节点的直接前驱     phead指向直接后驱*/ cout<<"输入ch的值: "<<endl;     gets(num);       /* 建立单链表"over"作为输入终止符*/    while(strcmp(num,"over")!=0) {  ip=new note2;  strcpy(ip->ch,num);
      if(head==NULL)   head=ip;  else    end->next=ip;  end=ip;  last++;  cout<<"输入ch的值:"<<endl;  gets(num); } if(end!=NULL)  end->next=NULL; if(head==NULL) {  cout<<"链表为空"<<endl;  return; } if(last<=2)                  //单链表反转           {   iq=end;  end->next=head;  end=head;  head=iq;  end->next=NULL; } else {  phead=head;  ir=phead->next;  pnext=ir->next;  while(pnext->next!=NULL)                       {   ir->next=phead;   phead=ir;   ir=pnext;   pnext=pnext->next;  }  ir->next=phead;  pnext->next=ir;  end=head;  head=pnext;  end->next=NULL; } cout<<"输出结果:"<<endl;                       //链表输出 ir=head; while(ir->next!=NULL) {  cout<<setw(4)<<ir->ch;  ir=ir->next; } cout<<setw(4)<<end->ch<<endl; cout<<endl;}
              
     
     

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  7. Posted by my-space on 2006年04月1日 at 14:53

    /*作业6.21顺序表完整程序*/#include<iostream.h>#include<stdio.h>#include<string.h>#include<iomanip.h>const int m=9;struct note{ char ch[10];};note array[m];void main(){ cout<<"输入9个字符串如e1,e2,….e9"<<endl; for(int i=0;i<m;i++)     //建立顺序表 {  cout<<"输入第"<<i+1<<"个字符串"<<endl;  gets(array[i].ch);} char nam[3]; if(m%2==0)          //如果m为偶数时作反转 {  for(int j=0;j<m/2;j++)  {   strcpy(nam,array[j].ch);   strcpy(array[j].ch,array[m-j-1].ch);   strcpy(array[m-j-1].ch,nam);  } } else         //如果m为奇数时作反转 {  for(int k=0;k<(m-1)/2;k++)  {   strcpy(nam,array[k].ch);   strcpy(array[k].ch,array[m-k-1].ch);   strcpy(array[m-k-1].ch,nam);  } }      cout<<"输出结果:"<<endl;           //输出最终结果 for(int x=0;x<m;x++)    cout<<setw(4)<<array[x].ch; cout<<endl;}  

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  8. Posted by my-space on 2006年04月4日 at 13:50

    //作业7-18
    #include<iostream.h>#include<stdio.h>#include<iomanip.h>#include<string.h>const  int maxsize=100;char *NEW();void DELETE(char*head,int i,int j);char array[maxsize];void main(){ int i,j; char *head; cout<<"输入i和j的值"<<endl;    cin>>i>>j; head=NEW();    cout<<"输出删除前的串"<<endl;    cout<<head<<endl; DELETE(head,i,j); cout<<"输出删除后的串"<<endl;    cout<<head<<endl;}//建表函数定义char *NEW(){ char *pi; cout<<"输入串"<<endl; pi=gets(array); return pi;}//删除函数定义void DELETE(char *head,int i,int j){ for(int k=i;head[k+j-1]!=”;k++)  head[k-1]=head[k+j-1]; head[k-1]=”;}
     

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  9. Posted by my-space on 2006年04月4日 at 13:53

    //作业7-20
    #include<iostream.h>#include<conio.h>typedef struct node{ char ch; node *next;}node;node *NEW();void RESULT(node *head);//mian函数定义void  main(){ node *head1,*pi; head1=NEW(); cout<<"输出未替代前的链表"<<endl; pi=head1->next; while(pi!=NULL) {  cout<<pi->ch;  pi=pi->next; } cout<<endl; RESULT(head1); cout<<"输出替代后的链表"<<endl; pi=head1->next; while(pi!=NULL) {  cout<<pi->ch;  pi=pi->next; } cout<<endl; return;
    }//下面建表node *NEW(){ node *head,*end,*ip; char num; head=new node;        //建立头节点并置空 end=head; cout<<"输入字符值以回车为结束符"<<endl; num=getche(); cout<<endl; while(num!=’\r’) {  ip=new node;  ip->ch=num;  end->next=ip;  end=ip;  cout<<"输入字符值以回车为结束符"<<endl;  num=getche();  cout<<endl; } end->next=NULL; return head;}//下面替代处理void RESULT(node *head){ node *ir; ir=head->next; while(ir!=NULL) {  if(ir->ch==’c’)   ir->ch=’s’;  ir=ir->next; } }
     
      
     

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